Decay properties for the Sobolev spaces

Question

Suppose we have a $f$ belonging to the Sobolev space $W^{1,1}(0,\infty)$, i.e., on $\mathbb R^{+}$. Then why does $f(x)$ decay to zero as $x$ goes to infinity?

Answer 1

First, the smooth functions with bounded support are dense in $W^{1,1}(0,\infty)$. I assume you already know that smooth functions are dense in $W^{1,1}(0,\infty)$; if not, I am sure this has been discussed on this site more than once. Let $f\in W^{1,1}(0,\infty)\cap C^\infty(0,\infty)$ and $\eta_K\in C^\infty([0,2K])$ with $0\leq \eta_K\leq 1$, $\eta_K(x)=1$ if $x\leq K$ and $|\eta_K'|\leq 2/K$. Clearly $f\eta_K\to f$ in $L^1(0,\infty)$. Moreover, $$ \|f'-(f\eta_K)'\|_{L^1}\leq \|f'-f'\eta_K\|_{L^1}+\frac 2 K\int_K^{2K}|f|\leq \|f'-f'\eta_K\|_{L^1}+\frac 2 K\|f\|_{L^1} $$ also converges to zero. Thus $f\eta_K\to f$ in $W^{1,1}(0,\infty)$.

Furthermore, if $f\in C^\infty(0,\infty)$ has support in $[0,K]$, then $$ |f(x)|=\left|f(2K)+\int_{2K}^x f'(t)\,dt\right|\leq \int_x^\infty |f'(t)|\,dt\leq \|f'\|_{W^{1,1}}. $$ Hence we can extend the inclusion $C_c^\infty([0,\infty))\to C_0([0,\infty))$ to a continuous operator from $W^{1,1}(0,\infty)\to C_0([0,\infty))$. Since convergence in $W^{1,1}$ implies a.e. convergence along a subsequence, this means in particular that every function in $W^{1,1}(0,\infty)$ has a continuous representative that vanishes at infinity.